The value of tan ^{-1}(-sqrt{3})-sin ^{-1}lef | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We know the principal value branches of the inverse trigonometric functions: $	an ^{-1}(-x) = -	an ^{-1}(x)$ $\sin ^{-1}\left(\frac{1}{\sqrt{2}}

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(C) We know the principal value branches of the inverse trigonometric functions: $	an ^{-1}(-x) = -	an ^{-1}(x)$ $\sin ^{-1}\left(\frac{1}{\sqrt{2}}ight) = \frac{\pi}{4}$ $\cos ^{-1}(-x) = \pi - \cos ^{-1}(x)$ Substituting these values into the expression: $	an ^{-1}(-\sqrt{3}) - \sin ^{-1}\left(\frac{1}{\sqrt{2}}ight) + \cos ^{-1}\left(-\frac{1}{2}ight)$ $= -	an ^{-1}(\sqrt{3}) - \frac{\pi}{4} + \left(\pi - \cos ^{-1}\left(\frac{1}{2}ight)ight)$ $= -\frac{\pi}{3} - \frac{\pi}{4} + \pi - \frac{\pi}{3}$ $= \pi - \left(\frac{\pi}{3} + \frac{\pi}{3} + \frac{\pi}{4}ight)$ $= \pi - \left(\frac{4\pi + 4\pi + 3\pi}{12}ight)$ $= \pi - \frac{11\pi}{12} = \frac{\pi}{12}$

The value of $\tan ^{-1}(-\sqrt{3})-\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\cos ^{-1}\left(\frac{-1}{2}\right)$ is

Similar Questions

If $\sin ^{-1} x=y,$ then

The value of $\cos ^{-1}\left(\tan \left(\frac{7 \pi}{4}\right)\right)$ is

$\tan ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+\sec ^{-1}\left(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}}\right)$ is equal to $.........$

${\left[ {\sin \left( {{{\tan }^{ - 1}}\frac{3}{4}} \right)} \right]^2} = $

$\sin (2{\sin ^{ - 1}}0.8) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module